
3.2 moles of hydrogen iodide were heated in a sealed bulb at ${{444}^{o}}C$ till the equilibrium state was reached. Its degree of dissociation at this temperature was found to be 22%. The number of moles of hydrogen iodide present at equilibrium is:
(A) 2.496
(B) 1.87
(C) 2
(D) 4
Answer
163.5k+ views
Hint: Chemical equilibrium is reached when a reversible reaction proceeds both backwards and forwards by the same amount at the same time. Its nature is dynamic. The phenomenon of forming free ions carrying current, which at a specific concentration is dissociated from the proportion of solute, is known as the degree of dissociation. The amount of the dissociated material, which can be expressed as the number of molecules or moles, can be divided by the total amount of the substance to determine the degree of dissociation.
Complete step by step solution:
Initially, 3.2 moles of hydrogen iodide were heated in a sealed bulb at ${{444}^{o}}C$ till the equilibrium state was reached. It is given that the degree of dissociation is 22%, that is, 22% of the molecules of hydrogen iodide are dissociated.
$\alpha =\frac{22}{100}\times 3.2$
$\alpha =0.704$
Initial concentration 3.2 0 0
At equilibrium $3.2-\alpha $ $\frac{\alpha }{2}$ $\frac{\alpha }{2}$
Hence, at equilibrium, the number of moles of hydrogen iodide present is:
= $3.2-\alpha $
= $3.2-0.704$
= $2.496$
Correct Option: (A) 2.496.
Note: To approach these types of numerical problems, one must focus on writing an appropriate equilibrium constant equation according to the given equilibrium equation and also take care of stoichiometric coefficients of reactants and products. These will help to calculate the appropriate equilibrium constant value without any mistake.
Complete step by step solution:
Initially, 3.2 moles of hydrogen iodide were heated in a sealed bulb at ${{444}^{o}}C$ till the equilibrium state was reached. It is given that the degree of dissociation is 22%, that is, 22% of the molecules of hydrogen iodide are dissociated.
$\alpha =\frac{22}{100}\times 3.2$
$\alpha =0.704$
Initial concentration 3.2 0 0
At equilibrium $3.2-\alpha $ $\frac{\alpha }{2}$ $\frac{\alpha }{2}$
Hence, at equilibrium, the number of moles of hydrogen iodide present is:
= $3.2-\alpha $
= $3.2-0.704$
= $2.496$
Correct Option: (A) 2.496.
Note: To approach these types of numerical problems, one must focus on writing an appropriate equilibrium constant equation according to the given equilibrium equation and also take care of stoichiometric coefficients of reactants and products. These will help to calculate the appropriate equilibrium constant value without any mistake.
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